AP Physics B - Multiple Choice Questions on Fluid Mechanics

Essential things you need to remember in fluid mechanics were discussed in the post dated 3^rd December 2007. A few multiple choice questions (with solution) in this section were discussed in the posts dated 4^th and 5^th December 2007. You can access all posts related to fluid mechanics on this site by clicking on the label ‘fluid mechanics’ below this post.

Today we will discuss a few more multiple choice questions on fluid mechanics:

(1) A water storage tank has a square hole of side 2 cm at its bottom. A plumber, unaware of the hole, admits water into the tank at a constant rate of √2 litre per second. Up to what height can water remain in the tank? (g = 10 ms^–2)

(a) 12.25 m

(b) 8.25 m

(c) 2.45 m

(d) 1.25 m

(e) 0.625 m

If h is the height up to which water can remain in the tank, the velocity v of water flowing out through the hole in the steady state is given by

v = √(2gh)

The volume of water flowing out in the steady state is av where a is the area of the hole.

This must be equal to the volume of incoming water which is given as √2 litre = √2 ×10^–3

m³. Thus we have

a√(2gh) = √2 ×10^–3

Here a = 4 cm² = 4×10^–4 m².

Substituting, 4×10^–4×√(2×10×h) = √2 ×10^–3

Or, √(10h) = 2.5 from which h = 6.25/10 = 0.625 m.

(2) A piece of granite floats at the interface of mercury and water contained in a beaker (Fig.). If the densities of granite, water and mercury are ρ, ρ₁ and ρ₂ respectively, the ratio of the volume of granite in water to the volume in mercury is

(a) (ρ₂ – ρ) /(ρ – ρ₁)

(b) (ρ₂ + ρ) /(ρ+ ρ₁)

(c) ρ₁ ρ₂ /ρ

(d) ρ₁ /ρ₂

(e) ρ₂ /ρ₁

The weight of a floating body is equal to the weight of the displaced fluid. If V and v represent the total volume of the piece of granite and volume of granite in water respectively, we have

V ρg = v ρ₁g + (V – v) ρ₂g

Or, v(ρ₁ – ρ₂) = V(ρ – ρ₂)

Therefore, v/V = (ρ – ρ₂) /(ρ₁ – ρ₂)

The ratio required in the question is v/ (V–v) and is given by

v/ (V–v) = (ρ – ρ₂) /[(ρ₁ – ρ₂) –(ρ – ρ₂)]

Or, v/ (V–v) = (ρ – ρ₂) /(ρ₁ – ρ) = (ρ₂ – ρ) /(ρ – ρ₁)

(3) The pressure of water at the bottom of a water tank exceeds the atmospheric pressure by 10⁴ pascal. The velocity of efflux of water through an orifice at the bottom of the water tank will be (g = 10 ms^–2)

(a) √5 ms^–1

(b) √10 ms^–1

(c) √15 ms^–1

(d) √20 ms^–1

(e) √30 ms^–1

If h is the height of water column in the tank, we have

hρg = 10⁴

Since the density (ρ) of water is 1000 kg m^–3,

h×1000×10 = 10⁴ from which h = 1m.

The velocity of efflux (Torricelli’s theorem) is √(2gh) = √(2×10×1) = √20 ms^–1

(4) In a wind tunnel the flow speeds (of air) on the upper and lower surfaces of the wing of a model airplane are v₁ and v₂ respectively (v₁> v₂). If the wing area is A and the density of air is ρ, the lift on the wing is

(a) ρ (v₁² – v₂²)A

(b) ½ ρ (v₁ – v₂)A

(c) ½ ρ (v₁² – v₂²)A

(d) ½ (v₁² – v₂²)A /ρ

(e) ρ (v₁ – v₂)A

If P₁ and P₂ are the pressures on the upper and lower surfaces of the wing we have, by Bernoulli’s theorem,

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

[We have ignored the small height difference between the top and bottom of the wing so that the gravitational potential energy is treated as constant].

The pressure at the top is less than the pressure at the bottom since v₁> v₂.

The lift on the wing is (P₂ – P₁)A = ½ ρ (v₁² – v₂²)A

You will find similar useful multiple choice questions with solution here as well as here at Physicsplus.

Fluid Mechanics- Multiple Choice Practice Questions on Force of Buoyancy

In the post dated 3^rd December 2007, the essential formulas you have to remember in the section 'Fluid Mechanics' were discussed. As promised, I will give you a few multiple choice questions with solution. Here is a question on force of buoyancy:

In the adjoining figure, a metallic block of volume 20 cm³ is shown suspended, using a thin nylon string, from an independent support. The entire block is immersed in water in a beaker, without touching the sides of the beaker. The reading of the electronic balance (on which the beaker is placed) is 257 gram, as shown.

If the weight of the beaker is 37 gram, what is the weight of the water in the beaker?

(a) 200 g (b) 220 g (c) 230 g (d) 240 g (e) Data insufficient for calculation

Since the metallic block is completely immersed in water, the force of buoyancy on the block is equal to the weight of 20 cm³ of water displaced by the block. In other words, water exerts an upthrust equal to the weight of 20 g of water. The block exerts an equal force downwards (reaction force) which the electronic balance registers in addition to the weight of the beaker and water.

The weight of the beaker and water together is therefore 257–20 = 237 g. Since the weight of the beaker is 37 g, the weight of water alone is 237–37 = 200 g.

Now, suppose the block was kept immersed using a rigid support instead of the string. Then also, the same explanation holds good. If you dip your finger gently in the water in the beaker, the reading of the balance will increase since the water exerts a force of buoyancy on your finger and your finger exerts an equal and opposite force of reaction.

Here is another multiple choice question (MCQ):

An iceberg floating in sea water just sinks when a mass of 360 kg is placed on it. What is the mass of the iceberg? (Relative density of sea water = 1.02; relative density of ice = 0.9)

(a) 1500 kg (b) 1800 kg (c) 2400 kg (d) 2700 kg (e) 3200 kg

The density of ice is 900 kg m^–3 and that of sea water is 1020 kg m^–3, as obtained from their relative densities. When the iceberg just sinks, the mass of displaced sea water is 1200V kg. where V is the volume of the iceberg. The mass of iceberg is 900V kg. Therefore we have, in this case (since the weight of floating body = weight of displaced liquid),

900V + 360 = 1020V, from which V = 3 m³

Therefore, mass of iceberg = 900×3 = 2700 kg.

Let us consider one more question:

A solid sphere of volume V remaining submerged in a liquid of density σ is lifted through a height ‘x’ within the liquid. If the density of the material of the sphere is ρ where ρ > σ, which one of the following statements is correct?

(a) There is no change in the potential energy of the sphere

(b) The potential energy of the sphere decreases by Vρgx

(c) The potential energy of the sphere increases by Vρgx

(d) The potential energy of the sphere increases by V[(ρ+σ)/2]gx

(e) The potential energy of the sphere increases by V(ρ – σ)gx

You will have to exert a force equal to the apparent weight of the sphere for lifting it. The apparent weight of the sphere is Vρg – Vσg = V(ρ – σ)g. The first term is its real weight and the second term is the up thrust (force of buoyancy) on it.

The work done in lifting the sphere through the height ‘x’ is V(ρ – σ)gx and this is the increase in potential energy of the sphere.

More questions from Fluid Mechanics will be discussed in the next post.

Meanwhile see a useful post in this section here.

Fluid Mechanics – Equations to be remembered

“There is no substitute for hard work.”

– Thomas A. Edison

You won’t be allowed to use equation tables and calculators to work out multiple choice questions (MCQ). (These are allowed in the case of free-response questions). You will have to remember certain basic formulas for solving multiple choice questions within the permitted time. The essential things you must remember in the section, ‘Fluid Mechanics’ are given below:

1. Weight of floating body = weight of displaced liquid

This can be modified as mass of floating body = mass of displaced fluid

2. Force of buoyancy (F_buoy)= weight of displaced fluid

or, F_buoy = Vρg where V is the volume of fluid displaced, ρ is the density of the fluid and g is the acceleration due to gravity.

3. Equation of continuity in fluid dynamics, as applied to liquids which are supposed to be incompressible, is A₁v₁= A₂v₂ where A₁ and A₂ are the cross section areas of the tube and v₁ and v₂ are the velocities of the liquid at these sections respectively.

4. Bernoulli’s equation (Bernoulli’s theorem) is

P + ρgh + (½)ρv² = constant where P is the pressure, ρ is the density, ‘h’ is the height (with respect to the reference level for estimating the gravitational potential energy) and ‘v’ is the velocity of the fluid. The symbol ‘y’ also may be used in place of ‘h’.

Note that Bernoulli’s theorem follows from the law of conservation of energy in the case of a small mass ‘m’ of a fluid throughout its flow and can be written as

PV+ mgh + (½ )mv² = constant

Another useful form of Bernoulli’s equation is

P/ρ + gh + (½)v² = constant

In the case of a liquid flowing through a horizontal pipe (constant gravitational potential energy), the pressure of liquid will be smaller at points where the velocity is greater and vice versa.

5.Torricelli’s Theorem:

Velocity of efflux (of liquid flowing out through a hole) =√(2gh)

where ‘h’ is the depth of the hole (fig).

(It is interesting to note that a particle dropped from a height ‘h’ will strike the ground with the above velocity)

The expression for the efflux velocity follows from Bernoulli’s theorem by considering the cases at the free surface of the liquid in the tank (fig) and at the hole:

P + ρgH + 0 = P + ρg(H–h) + (½)ρv² so that v =√(2gh),

where P is the atmospheric pressure. Note that the hole is open to the atmosphere, as is the free liquid surface in the tank.

6. The horizontal range of liquid jet (fig), R= 2√[h(H–h)]

Note that the above expression is obtained by considering the horizontal range (on the ground) of a particle projected horizontally from a height (H–h). Do it as an exercise.

The range R will be maximum if h = H/2. [You may show this by putting dR/dh = 0 when R is maximum]. The maximum range is H.

In the next post we will consider some typical questions in this section. Of course, their solution also will be discussed.